Resposta :

Relações:

 

[tex]cossec(\theta)=\frac{1}{sen(\theta)}[/tex]

 

[tex]cotg(\theta)=\frac{1}{tg(\theta)}[/tex]

 

[tex]\pi=180[/tex]

 

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[tex]a)\frac{1}{sen(\frac{3\pi}{4})}=\frac{1}{sen(135)}=\boxed{\frac{2}{\sqrt2}}[/tex]

 

[tex]b)sen(195)=sen(15)=\boxed{-0,259}[/tex]

 

[tex]c)\frac{1}{tg(\frac{5\pi}{3})}=\frac{1}{sen(300)}=\boxed{-\frac{2}{\sqrt3}}[/tex]

 

Celio

Olá, Cibele.

 

[tex]a)\ \csc \frac34\pi=\underbrace{\frac1{\sin(\pi-\frac{\pi}4)}=\frac1{\sin \pi \cdot \cos\frac{\pi}4 - \sin \frac{\pi}4 \cdot \cos\pi}}_{\sin (a - b) =\sin a \cdot \cos b - \sin b \cdot \cos a}=\\\\=\frac{1}{0-\frac{\sqrt2}2\cdot(-1)}=\frac2{\sqrt2}=\frac{2\sqrt2}{\sqrt2 \cdot \sqrt2}=\boxed{\sqrt2}[/tex]

 

 

[tex]b)\ \sin 195\°=\sin(180\º+15\º)=\\\\=\underbrace{\sin180\º\cos15\º+\sin15\º\cos180\º}_{\sin (a + b) = \sin a \cdot \cos b + \sin b \cdot \cos a}=\\\\=0-\sin15\º=-\sin(45\º-30\º)=\\\\= -(\underbrace{\sin45\º\cos30\º-\sin30\º\cos45\º}_{\sin (a - b) = \sin a \cdot \cos b - \sin b \cdot \cos a})=-(\frac{\sqrt2}2\cdot \frac{\sqrt3}2-\frac12\cdot \frac{\sqrt2}2)=\\\\\\=-(\frac{\sqrt6}4-\frac{\sqrt2}4)=\boxed{\frac{\sqrt2-\sqrt6}4}[/tex]

 

 

[tex]c)\ \cot \frac53 \pi=\frac1{\tan(2\pi-\frac{\pi}3)}=\frac{\cos(2\pi-\frac{\pi}3)}{\sin(2\pi-\frac{\pi}3)}=\frac{\cos2\pi\cos\frac{\pi}3-\sin2\pi\sin\frac{\pi}3}{\sin2\pi\cos\frac{\pi}3-\sin\frac{\pi}3\cos2\pi}=\\\\ =\frac{-\frac12-0}{0+\frac{\sqrt3}2}=-\frac12 \cdot \frac2{\sqrt3}=-\frac1{\sqrt3}\cdot\frac{\sqrt3}{\sqrt3}=\boxed{-\frac{\sqrt3}3}[/tex]

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